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Exercise 4–1
Ex: 4.1 Refer to Fig. 4.3(a). For v I ≥ 0, the diode conducts and presents a zero voltage drop. Thus v O = v I . For v I < 0, the diode is cut off, zero current flows through R, and v O = 0. The result is the transfer characteristic in Fig. E4.1.
(c)
V5V
I0A Ex: 4.2 See Fig. 4.3a and 4.3b. During the positive half of the sinusoid, the diode is forward biased, so it conducts resulting in v D = 0. During the negative half cycle of the input signal v I , the diode is reverse biased. The diode does not conduct, resulting in no current flowing in the circuit. So v O = 0 and v D = v I − v O = v I . This results in the waveform shown in Fig. E4.2.
2.5 k
5 V (d)
V0V
10 V vˆI = = 10 mA Ex: 4.3 ˆiD = R 1 k 1 dc component of v O = vˆO π 1 10 = vˆI = π π = 3.18 V
05 2.5 2 mA
I
2.5 k
5 V (e)
I 3V
Ex: 4.4
0 V3V
2V
(a)
0 1V
5 V
I
2.5 k
50 2 mA 2.5 (f)
3 3 mA 1
I
1 k
5 V
V0V
1 k 0 3 V
(b)
0
2 V
5 V
1 V 2.5 k
I0A V5V
51 1 4 mA
I
V1V
I
Ex: 4.5 Vavg =
10 π
10 10 k 50 + R = π = 1 mA π ∴ R = 3.133 k
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Exercise 4–2
R 10 k
Ex: 4.6 Equation (4.5) I2 V2 − V1 = 2.3 V T log I1 At room temperature VT = 25 mV
VCC 5 V
10 V2 − V1 = 2.3 × 25 × 10−3 × log 0.1
VD
I2 V2 − V1 = 2.3 × VT log I1 I2 V2 = V1 + 2.3 × VT log I1
= 115 mV Ex: 4.7 i = IS ev /VT
(1)
First iteration
1 (mA) = IS e
(2)
V2 = 0.7 + 2.3 × 25 × 10−3 log
0.7/VT
Dividing (1) by (2), we obtain i (mA) = e
ID
(v −0.7)/VT
⇒ v = 0.7 + 0.025 ln(i) where i is in mA. Thus, for i = 0.1 mA,
= 0.679 V Second iteration 5 − 0.679 = 0.432 mA I2 = 10 k
0.43 1
V2 = 0.7 + 2.3 × 25.3 × 10−3 log
v = 0.7 + 0.025 ln(0.1) = 0.64 V
= 0.679 V 0.68 V
and for i = 10 mA,
we get almost the same voltage.
v = 0.7 + 0.025 ln(10) = 0.76 V
0.432 1
ID = 0.43 mA, VD = 0.68 V b. Use constant voltage drop model:
IS = 10−14 × 1.15T
VD = 0.7 V
= 1.17 × 10−8 A
ID = 1V = 1 μA 1 M
∴ The iteration yields
Ex: 4.8 T = 125 − 25 = 100◦ C
Ex: 4.9 At 20◦ C I =
constant voltage drop
5 − 0.7 = 0.43 mA 10 k
Ex: 4.11
Since the reverse leakage current doubles for every 10◦ C increase, at 40◦ C
10 V
I
R
I = 4 × 1 μA = 4 μA
⇒ V = 4 μA × 1 M = 4.0 V @ 0◦ C
I=
1 μA 4
2.4 V
1 ⇒ V = × 1 = 0.25 V 4
Ex: 4.10 a. Use iteration: Diode has 0.7 V drop at 1 mA current. Assume VD = 0.7 V ID =
5 − 0.7 = 0.43 mA 10 k
Diodes have 0.7 V drop at 1 mA ∴ 1 mA = IS e0.7/VT At a current I (mA), I = IS eVD /VT
Use Eq. (4.5) and note that
Using (1) and (2), we obtain
V1 = 0.7 V,
I = e(VD −0.7)/VT
I1 = 1 mA
(1)
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(2)
Exercise 4–3
For an output voltage of 2.4 V, the voltage drop 2.4 = 0.8 V across each diode = 3 Now I, the current through each diode, is
(d)
V 0.7 V
I = e(0.8−0.7)/0.025 = 54.6 mA 10 − 2.4 R= 54.6 × 10−3
0.7 5 2.5 1.72 mA
I
2.5 k
= 139
5 V (e)
Ex: 4.12
I 3V
(a)
0 5 V
V 3 0.7 2.3 V
2V 0 1V
2.5 k
5 0.7 I 1.72 mA 2.5
2.3 1 2.3 mA
I
V 0.7 V
1 k
(f)
5V
5 1.7 1 3.3 mA
I (b)
5 V
3 V 2 V
I0A
2.5 k
1 V
1 k V 1 0.7
0
1.7 V
0 I
V5V
(c)
Ex: 4.13 rd =
VT ID
ID = 0.1 mA
rd =
25 × 10−3 = 250 0.1 × 10−3
ID = 1 mA
rd =
25 × 10−3 = 25 1 × 10−3
ID = 10 mA
rd =
25 × 10−3 = 2.5 10 × 10−3
V5V
Ex: 4.14 For small signal model, iD = v D /rd
2.5 k
I0A
VT where rd = ID For exponential model,
5 V
iD = IS eV /V T
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(1)
Exercise 4–4
iD2 = e(V2 −V1 ) /V T = ev D /V T iD1 iD = iD2 − iD1 = iD1 ev D /V T − iD1 = iD1 ev D /V T − 1
c. If iD = 5 − i L = 5 − 1 = 4 mA.
(2)
In this problem, iD1 = ID = 1 mA. Using Eqs. (1) and (2) with VT = 25 mV, we obtain v D (mV)
a b c d
− 10 −5 +5 + 10
iD (mA) small signal
iD (mA) exponential model
− 0.4 − 0.2 + 0.2 + 0.4
− 0.33 − 0.18 + 0.22 + 0.49
Across each diode the voltage drop is ID VD = VT ln IS 4 × 10−3 = 25 × 10−3 × ln 4.7 × 10−16 = 0.7443 V Voltage drop across 4 diodes = 4 × 0.7443 = 2.977 V so change in VO = 3 − 2.977 = 23 mV.
Ex: 4.16 For a zener diode VZ = VZ0 + IZ rZ 10 = VZ0 + 0.01 × 50 VZ0 = 9.5 V
Ex: 4.15
For IZ = 5 mA,
15 V
VZ = 9.5 + 0.005 × 50 = 9.75 V For IZ = 20 mA,
R IL VO
VZ = 9.5 + 0.02 × 50 = 10.5 V
Ex: 4.17
15 V
R
I 5.6 V
VO 20 mV a. In this problem, = = 20 . iL 1 mA ∴ Total small-signal resistance of the four diodes = 20 20 ∴ For each diode, rd = = 5 . 4 VT 25 mV ⇒5= . But rd = ID ID ∴ ID = 5 mA 15 − 3 = 2.4 k. 5 mA b. For VO = 3 V, voltage drop across each diode 3 = = 0.75 V 4 and R =
iD = IS eV /VT IS =
iD eV /VT
=
5 × 10−3 = 4.7 × 10−16 A e0.75/0.025
0 to 15 mA
The minimum zener current should be 5 × IZk = 5 × 1 = 5 mA Since the load current can be as large as 15 mA, we should select R so that with IL = 15 mA, a zener current of 5 mA is available. Thus the current should be 20 mA, leading to R=
15 − 5.6 = 470 20 mA
Maximum power dissipated in the diode occurs when IL = 0 is Pmax = 20 × 10−3 × 5.6 = 112 mV
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Exercise 4–5
a. The diode starts conduction at
Ex: 4.18
v S = VD = 0.7 V
15 V
√ v S = Vs sin ωt, here Vs = 12 2
I
200
15 5.1 49.5 mA 0.2 k 50 mA
At ωt = θ,
VZ
v S = Vs sin θ = VD = 0.7 V √ 12 2 sin θ = 0.7 0.7 θ = sin−1 2.4◦ √ 12 2
Conduction starts at θ and stops at 180 − θ.
Thus, at no load VZ
5.1 V
∴ Total conduction angle = 180 − 2θ = 175.2◦ For line regulation: b. v O,avg
vo 200 vi
mV vo 7 = 33.8 = vi 200 + 7 V
200 VO IL
rZ
V O −IL (rZ 200 ) = IL IL mA = −6.8
(Vs sin φ − VD ) d φ θ
1 [−Vs cos φ − VD φ]φ=π−θ φ−θ 2π 1 [Vs cos θ − Vs cos (π − θ) − VD (π − 2θ )] = 2π But cos θ 1, cos (π − θ) − 1, and π − 2θ π VD 2Vs − 2π 2 VD Vs − = π 2 √ For Vs = 12 2 and VD = 0.7 V √ 12 2 0.7 v O,avg = − = 5.05 V π 2 c. The peak diode current occurs at the peak diode voltage. √ 12 2 − 0.7 V − VD ∴ ˆiD = s = R 100 = 163 mA √ PIV = +V S = 12 2
v O,avg =
For load regulation:
VZ0
(π−θ )
=
7
Line regulation =
1 = 2π
17 V
mV mA
Ex: 4.20
vS input
Ex: 4.19
Vs
vS Vs 12 2
VD 0
VD
0 u
u
2
t
output ( )
VS a. As shown in the diagram, the output is zero between (π − θ) to (π + θ) = 2θ
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t
Exercise 4–6
Here θ is the angle at which the input signal reaches VD .
Vs
∴ Vs sin θ = VD VD θ = sin−1 Vs VD 2θ = 2 sin−1 Vs
Vs (a) VO,avg =
1 2π
(Vs sin φ − 2VD ) d φ
2 [−Vs cos φ − 2VD φ]π−θ φ=θ 2π 1 = [Vs cos φ − Vs cos(π − θ) − 2VD (π − 2θ )] π But cos θ ≈ 1, π . 2
Peak current Vs − VD Vs sin (π /2) − VD = = R R If v S is 12 V(rms), √ √ then Vs = 2 × 12 = 12 2 √ 12 2 − 0.7 Peak current = 163 mA 100 Nonzero output occurs for angle = 2 (π − 2θ) The fraction of the cycle for which v O > 0 is 2 (π − 2θ) × 100 2π 0.7 2 π − 2 sin−1 √ 12 2 × 100 = 2π =
cos (π − θ) ≈ − 1 π − 2θ ≈ π. Thus ⇒ VO,avg
2Vs − 2VD π
√ 2 × 12 2 − 1.4 = 9.4 V π Peak voltage (b) Peak diode current = R √ Vs − 2VD 12 2 − 1.4 = = R 100 = 156 mA √ PIV = Vs − VD = 12 2 − 0.7 = 16.3 V =
Ex: 4.22 Full-wave peak rectifier:
D1 vO
vS vS
97.4 % Average output voltage VO is √ Vs 2 × 12 2 VO = 2 − VD = − 0.7 = 10.1 V π π Peak diode current ˆiD is √ ˆiD = Vs − VD = 12 2 − 0.7 R 100
D
S
=
for θ small.
c. Peak current occurs when φ =
2VV
sin–1
1 −θ [−Vs cos φ − VD φ]πφ=θ π Vs − VD , π
input t
0
θ
2
output
2VD
b. Average value of the output signal is given by ⎡ ⎤ (π −θ) 1 ⎣ 2× VO = (Vs sin φ − VD ) d φ ⎦ 2π =
Ex: 4.21
R
C
D2 Vp
t
Vr{
assume ideal diodes t
= 163 mA PIV = Vs − VD + VS √ √ = 12 2 − 0.7 + 12 2 = 33.2 V
T 2
The ripple voltage is the amount of voltage reduction during capacitor discharge that occurs
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Exercise 4–7
when the diodes are not conducting. The output voltage is given by
If t = 0 is at the peak, the maximum diode current occurs at the onset of conduction or at t = −ωt.
v O = Vp e−t/RC
During conduction, the diode current is given by T /2
Vp − Vr = Vp e− RC
←
discharge is only half T the period. We also assumed t . 2
T /2 Vr = Vp 1 − e− RC T /2 , for CR T/2 RC T /2 Thus Vr Vp 1 − 1 + RC T /2
e− RC 1 −
Vr =
Vp 2fRC
(a)
Q.E.D.
QSUPPLIED = QLOST iCav t = CV r
SUB (a)
Vp Vp iD,av − IL t = C = 2fRC 2fR Vp π ωR
iD,av =
Vp π + IL ωtR
where ωt is the conduction angle. Note that the conduction angle has the same expression as for the half-wave rectifier and is given by Eq. (4.30), 2Vr ∼ ωt = (b) Vp Substituting for ωt, we get ⇒ iD,av =
πVp 2Vr ·R Vp
= IL 1 + π
d v S + iL dt t=−ωt
assuming iL is const. iL
Vp = IL R
d Vp cos ωt + IL dt = −C sin ω t × ωVp + IL =C
= −C sin(−ωt) × ωVp + IL For a small conduction angle
Vp 2Vr
⇒ iD,max = Cωt × ωVp + IL Sub (b) to get 2V r iD,max = C ωVp + IL Vp Substituting ω = 2πf and using (a) together with Vp /R IL results in Vp Q.E.D. iDmax = IL 1 + 2π 2Vr
Ex: 4.23
D4 ac line voltage
D1 vO
vS
R
D2
C
D3
+ IL
Since the output is approximately held at Vp , Vp ≈ IL · Thus R Vp ⇒ iD,av ∼ + IL = π IL 2Vr
iD,max = C
sin(−ωt) ≈ − ωt. Thus
To find the average diode current, note that the charge supplied to C during conduction is equal to the charge lost during discharge.
=
iD = iC + iL
Q.E.D.
The output voltage, v O , can be expressed as v O = Vp − 2VD e−t/RC At the end of the discharge interval v O = Vp − 2VD − Vr The discharge occurs almost over half of the time period T /2. For time constant RC
T 2
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Exercise 4–8
e−t/RC 1 −
1 T × 2 RC
Ex: 4.24
T 1 ∴ VP − 2VD − Vr = Vp − 2VD 1 − × 2 RC
⇒ Vr = Vp − 2VD
i vI
i
T × 2RC
iD
vO
√ Here Vp = 12 2 and Vr = 1 V
iR
VD = 0.8 V T=
1 k
1 1 = s f 60
√ 1 = (12 2 − 2 × 0.8) ×
1 2 × 60 × 100 × C
√ (12 2 − 1.6) C= = 1281 μF 2 × 60 × 100
Without considering the ripple voltage, the dc output voltage √ = 12 2 − 2 × 0.8 = 15.4 V If ripple voltage is included, the output voltage is √ Vr = 14.9 V = 12 2 − 2 × 0.8 − 2 IL =
vA
vD
14.9 0.15 A 100
The conduction angle ωt can be obtained using √ Eq. (4.30) but substituting Vp = 12 2 − 2 × 0.8: 2Vr 2×1 ωt = = √ Vp 12 2 − 2 × 0.8
The average and peak diode currents can be calculated using Eqs. (4.34) and (4.35): Vp 14.9 V iDav = IL 1 + π , where IL = , 2Vr 100 √ Vp = 12 2 − 2 × 0.8, and Vr = 1 V; thus
iDmax = I 1 + 2π
iD = IS ev D /VT iD = e(v D −0.7)/V T 1 mA iD + 0.7 V ⇒ v D = VT ln 1 mA For v I = 10 mV, v O = v I = 10 mV It is an ideal op amp, so i+ = i− = 0. 10 mV = 10 μA 1 k 10 μA + 0.7 = 0.58 V v D = 25 × 10−3 ln 1 mA
∴ iD = iR =
v A = v D + 10 mV = 0.58 + 0.01 = 0.59 V For v I = 1 V
= 0.36 rad = 20.7◦
iDav = 1.45 A
The diode has 0.7 V drop at 1 mA current.
Vp 2Vr
vO = vI = 1 V vO 1 = = 1 mA 1 k 1 k v D = 0.7 V
iD =
VA = 0.7 V + 1 k × 1 mA = 1.7 V For v I = −1 V, the diode is cut off. ∴ vO = 0 V v A = −12 V
= 2.76 A Ex: 4.25
PIV of the diodes = VS − VDO
√ = 12 2 − 0.8 = 16.2 V
To provide a safety margin, select a diode capable of a peak current of 3.5 to 4A and having a PIV rating of 20 V.
vI
vO R
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IL
Exercise 4–9
v I > 0 ∼ diode is cut off, loop is open, and the opamp is saturated: vO = 0 V v I < 0 ∼ diode conducts and closes the negative feedback loop: vO = vI
For v I ≤ −5 V, diode D1 conducts and 1 (+v I + 5) 2
vI V = −2.5 + 2 For v I ≥ 5 V, diode D2 conducts and
v O = −5 +
1 (v I − 5) 2
vI V = 2.5 + 2 v O = +5 +
Ex: 4.26
10 k
D2
D1 5V
vO
5V v O
vI
Ex: 4.27 Reversing the diode results in the peak output voltage being clamped at 0 V:
10 k
10 k
Both diodes are cut off
t 10 V
Here the dc component of v O = VO = −5 V
for −5V ≤ v I ≤ + 5V and v O = v I
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Chapter 4–1
4.3
4.1
1
(a)
I
1.5 V
VD
D1 1 V
(a)
Cutoff
D2
V2V
2 V Conducting
1
I
2 k
I
1.5 V
VD
2 (3) 2 k
2.5 mA
3 V
(b)
(b)
3 V
(a) The diode is reverse biased, thus I =0A
2 k Conducting
VD = −1.5 V
3 (1) 2 1 mA
I
D1
1 V
V1V
(b) The diode is forward biased, thus VD = 0 V
2 V D2
1.5 V = 1.5 A I= 1 4.2 Refer to Fig. P4.2. (a) Diode is conducting, thus
4.4 (a)
V = −3 V I=
Cutoff
vO
+3 − (−3) = 0.6 mA 10 k
5V
(b) Diode is reverse biased, thus
0
t
I =0 V = +3 V
V p+ = 5 V Vp− = 0 V f = 1 kHz
(c) Diode is conducting, thus
(b)
V = +3 V I=
vO
+3 − (−3) = 0.6 mA 10 k
(d) Diode is reverse biased, thus I =0 V = −3 V
0
t 5 V
Vp+ = 0 V Vp− = −5 V f = 1 kHz
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Chapter 4–2
(c)
vO 0 V
(h)
vO
t 0
t v O = 0 V ∼ The output is always shorted to ground as D1 conducts when v I > 0 and D2 conducts when v I < 0.
vO = 0 V
(i)
vO
Neither D1 nor D2 conducts, so there is no output.
5V
(d)
vO
t 2.5 V
5V t
Vp+ = 5 V, Vp− = −2.5 V, Vp+ = 5 V,
Vp− = 0 V,
f = 1 kHz
When v I > 0, D1 is cut off and v O follows v I .
f = 1 kHz
When v I < 0, D1 is conducting and the circuit becomes a voltage divider where the negative peak is
Both D1 and D2 conduct when v I > 0 (e)
vO
1 k × −5 V = −2.5 V 1 k + 1 k (j)
vO
5V 5 V Vp+ = 5 V,
Vp− = −5 V,
t
5V
f = 1 kHz
D1 conducts when v I > 0 and D2 conducts when v I < 0. Thus the output follows the input. (f)
2.5 V Vp+ = 5 V, Vp− = −2.5 V,
t
f = 1 kHz
When v I > 0, the output follows the input as D1 is conducting.
vO 5V t
When v I < 0, D1 is cut off and the circuit becomes a voltage divider. (k)
vO Vp+ = 5 V,
Vp− = 0 V,
f = 1 kHz
5V
D1 is cut off when v I < 0
1V t
(g)
vO
5 V
4 V
t 5 V
Vp+ = 1 V, Vp− = −4 V, f = 1 kH3 When v I > 0, D1 is cut off and D2 is conducting. The output becomes 1 V.
Vp+ = 0 V,
Vp− = −5 V,
f = 1 kHz
D1 shorts to ground when v I > 0 and is cut off when v I < 0 whereby the output follows v I .
When v I < 0, D1 is conducting and D2 is cut off. The output becomes: vO = vI + 1 V
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Chapter 4–3
X = AB, Y = A + B
4.5
X and Y are the same for A=B X and Y are opposite if A = B
4.7 The case for the highest current in a single diode is when only one input is high: VY = 5 V VY ≤ 0.2 mA ⇒ R ≥ 25 k R From Fig. P4.5 we see that when v I < VB ; that is, v I < 3 V, D1 will be conducting the current I and iB will be zero. When v I exceeds the battery voltage (3 V), D1 cuts off and D2 conducts, thus steering I into the battery. Thus, iB will have the waveform shown in the figure. Its peak value will be 60 mA. To obtain the average value, we first determine the conduction angle of D2 , (π − 2θ), where 3 = 30◦ θ = sin−1 6
4.8 The maximum input current occurs when one input is low and the other two are high. 5−0 ≤ 0.2 mA R R ≥ 25 k
4.9
3 V
Thus π − 2θ = 180◦ − 60 = 120◦
12 k
0.33 mA
The average value of iB will be 60 × 120◦ = 20 mA 360◦ If the peak value of v I is reduced by 10%, i.e. from 6 V to 5.4 V, the peak value of iB does not change. The conduction angle of D2 , however, changes since θ now becomes 3 = 33.75◦ θ = sin−1 5.4
1 V
iB |av =
I0
D1 OFF
D2 ON V 1 V
0.33 mA 6 k
and thus
3 V (a)
π − 2θ = 112.5◦ Thus the average value of iB becomes iB |av =
60 × 112.5◦ = 18.75 mA 360◦
4.6 A
B
X
Y
0 0 1 1
0 1 0 1
0 0 0 1
0 1 1 1
(a) If we assume that both D1 and D2 are conducting, then V = 0 V and the current in D2 will be [0 − (−3)]/6 = 0.5 mA. The current in the 12 k will be (3 − 0)/12 = 0.25 mA. A node equation at the common anodes node yields a negative current in D1 . It follows that our assumption is wrong and D1 must be off. Now making the assumption that D1 is off and D2 is on, we obtain the results shown in Fig. (a): I =0 V = −1 V
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Chapter 4–4
3 V
4.12 D
RS
6 k
3 – 0 0.5 mA 6
vI
0V D1 ON
0.25 mA
D2 ON
0 – (–3) 12 0.25 mA
vO R vI 1 vI R RS 2
iD
V0
vI R RS
D starts to conduct when v I > 0
12 k
vO
3 V (b)
vI
0
(b) In (b), the two resistors are interchanged. With some reasoning, we can see that the current supplied through the 6-k resistor will exceed that drawn through the 12-k resistor, leaving sufficient current to keep D1 conducting. Assuming that D1 and D2 are both conducting gives the results shown in Fig. (b):
4.13 For v I > 0 V: D is on, v O = v I , iD = v I /R For v I < 0 V: D is off , v O = 0, iD = 0
I = 0.25 mA V =0V 4.10 The analysis is shown on the circuit diagrams below. √ 120 2 ≥ 4.2 k 40 The largest reverse voltage appearing across the diode is equal to the peak input voltage: √ 120 2 = 169.7 V 4.11 R ≥
These figures belong to Problem 4.10.
reverse biased 10 10 5 k
5 k
5 k
V I
10 10 10 2.5 V
5
20 k
I
2.5 0.1 mA 5 20
I0
2.5 V
V
1.5 V
V 1.5 2.5 1 V (b)
V 0.1 20 2 V (a)
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Chapter 4–5
∴ Peak-to-peak sine wave voltage
4.14
= 2A = 34 V Given the average diode current to be 1 2π
2π
A sin φ − 12 d φ = 100 mA R
0
1 2π
−17 cos φ − 12φ R
φ = 0.75π = 0.1 φ = 0.25π
R = 8.3 A − 12 = 0.6 A R Peak reverse voltage = A + 12 = 29 V Peak diode current =
For resistors specified to only one significant digit and peak-to-peak voltage to the nearest volt, choose A = 17 so the peak-to-peak sine wave voltage = 34 V and R = 8 . Conduction starts at v I = A sin θ = 12 17 sin θ = 12 π rad θ= 4 Conduction stops at π − θ.
4.15
R vI
∴ Fraction of cycle that current flows is
D
12 V
vI A 12 V
Peak diode current 17 − 12 = 0.625 A = 8 Peak reverse voltage = A + 12 = 29 V
2
0
π − 2θ × 100 = 25% 2π Average diode current = 1 −17 cos φ − 12φ φ = 3π/4 = 103 mA 2π 8 φ = π/4
conduction occurs v I = A sin θ = 12 ∼ conduction through D occurs For a conduction angle (π − 2θ ) that is 25% of a cycle 1 π − 2θ = 2π 4 π θ= 4 A = 12/sin θ = 17 V
4.16 V RED 3V ON 0 OFF –3 V OFF
4.17 VT =
GREEN OFF - D1 conducts OFF - No current flows ON - D2 conducts
kT q
where k = 1.38 × 10−23 J/K = 8.62 × 10−5 eV/K T = 273 + x◦ C q = 1.60 × 10−19 C
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Chapter 4–6
Thus
I = 10−3 e(0.6 − 0.7)/0.025
VT = 8.62 × 10−5 × (273 × x◦ C), V
= 18.3 μA
x [◦ C]
VT [mV]
−55 0 +55 +125
18.8 23.5 28.3 34.3
To increase the current by a factor of 10, VD must be increased by VD , 10 = eVD /0.025 ⇒ VD = 0.025 ln10 = 57.6 mV 4.21 IS can be found by using IS = ID · e−VD /VT .
for VT = 25 mV at 17◦ C
Let a decrease by a factor of 10 in ID result in a decrease of VD by V: ID = IS eVD /VT
4.18 i = I S ev /0.025
ID = IS e(VD −V)/VT = IS eVD /VT · e−V /V T 10
∴ 10,000IS = IS ev /0.025
Taking the ratio of the above two equations, we have
v = 0.230 V At v = 0.7 V,
10 = eV /V T ⇒ V 60 mV
i = IS e0.7/0.025 = 1.45 × 1012 IS
Thus the result in each case is a decrease in the diode voltage by 60 mV. (a) VD = 0.700 V, ID = 1 A ⇒ IS = 6.91 × 10−13 A;
4.19 I1 = IS e0.7/VT = 10−3 i2 = IS e0.5/VT
10% of ID gives VD = 0.64 V
i2 i2 0.5 − 0.7 = −3 = e 0.025 i1 10
(b) VD = 0.650 V, ID = 1 mA ⇒ IS = 5.11 × 10−15 A;
i2 = 0.335 μA
10% of ID gives VD = 0.59 V (c) VD = 0.650 V, ID = 10 μA ⇒ IS = 5.11 × 10−17 A;
4.20 I = IS eVD /VT 10−3 = IS e0.7/VT
(1)
10% of ID gives VD = 0.59 V
(2)
(d) VD = 0.700 V, ID = 100 mA ⇒ IS = 6.91 × 10−14 A;
For VD = 0.71 V, I = IS e
0.71/VT
10% of ID gives VD = 0.64 V
Combining (1) and (2) gives I = 10−3 e(0.71 − 0.7)/0.025
4.22 IS can be found by using IS = ID · e−VD /VT .
= 1.49 mA For VD = 0.8 V, (3)
I = IS e0.8/VT Combining (1) and (3) gives I = 10−3 × e(0.8 − 0.7)/0.025 = 54.6 mA Similarly, for VD = 0.69 V we obtain −3
I = 10
×e
Let an increase by a factor of 10 in ID result in an increase of VD by V: ID = IS eVD /VT 10I D = IS e(VD +V )/VT = IS eVD /VT · eV /V T Taking the ratio of the above two equations, we have 10 = eV /V T ⇒ V 60 mV
(0.69 − 0.7)/0.025
= 0.67 mA
Thus the result is an increase in the diode voltage by 60 mV.
and for VD = 0.6 V we have
Similarly, at ID /10, VD is reduced by 60 mV.
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Chapter 4–7
(a) VD = 0.70 V, ID = 10 mA ⇒ IS = 6.91 × 10−15 A;
4.25
I
ID × 10 gives V D = 0.76 V ID /10 gives VD = 0.64 V (b) VD = 0.70 V, ID = 1 mA ⇒ IS = 6.91 × 10−16 A;
ID1
ID2
D1
D2
VD
ID × 10 gives V D = 0.76 V
ID /10 gives VD = 0.64 V (c) VD = 0.80 V, ID = 10 A ⇒ IS = 1.27 × 10−13 A; ID × 10 gives V D = 0.86 V ID /10 gives VD = 0.74 V (d) VD = 0.70 V, ID = 1 mA ⇒ IS = 6.91 × 10−16 A; ID × 10 gives V D = 0.76 V
ID1 = IS1 eVD /VT
(1)
VD /VT
(2)
ID2 = IS2 e
Summing (1) and (2) gives ID1 + ID2 = (IS1 + IS2 )eVD /VT But ID1 + ID2 = I Thus
ID /10 gives VD = 0.64 V
I = (IS1 + IS2 ) eVD /VT
(e) VD = 0.6 V, ID = 10 μA ⇒ IS = 3.78 × 10−16 A
From Eq. (3) we obtain I VD = VT ln IS1 + IS2
ID × 10 gives V D = 0.66 V ID /10 gives VD = 0.54 V
4.23 The voltage across three diodes in series is 2.0 V; thus the voltage across each diode must be 0.667 V. Using ID = IS eVD /VT , the required current I is found to be 3.9 mA.
(3)
Also, Eq. (3) can be written as IS2 I = IS1 eVD /VT 1 + IS1
(4)
Now using (1) and (4) gives ID1 =
I IS1 =I 1 + (IS2 /IS1 ) IS1 + IS2
We similarly obtain If 1 mA is drawn away from the circuit, ID will be 2.9 mA, which would give VD = 0.794 V, giving an output voltage of 1.98 V. The change in output voltage is −22 mV.
ID2 =
I IS2 =I 1 + (IS1 /IS2 ) IS1 + IS2
4.26 4.24 Connecting an identical diode in parallel would reduce the current in each diode by a factor of 2. Writing expressions for the currents, we have
I
ID = IS eVD /VT ID = IS e(VD −V)/VT = IS eVD /VT · e−V /V T 2 Taking the ratio of the above two equations, we have 2=e
V /V T
D1
D2
D3
I1
2I1
4I1
D4 8I1
⇒ V = 17.3 mV
Thus the result is a decrease in the diode voltage by 17.3 mV.
I1 0.1 mA
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Chapter 4–8
The junction areas of the four diodes must be related by the same ratios as their currents, thus A4 = 2A3 = 4A2 = 8 A1 With I1 = 0.1 mA, I = 0.1 + 0.2 + 0.4 + 0.8 = 1.5 mA 4.27 We can write a node equation at the anodes: ID2 = I1 − I2 = 7 mA ID1 = I2 = 3 mA We can write the following equation for the diode voltages: V = VD2 − VD1
Solving the above equation, we have R = 42 4.29 For a diode conducting a constant current, the diode voltage decreases by approximately 2 mV per increase of 1◦ C. T = −20◦ C corresponds to a temperature decrease of 40◦ C, which results in an increase of the diode voltage by 80 mV. Thus VD = 770 mV. T = + 85◦ C corresponds to a temperature increase of 65◦ C, which results in a decrease of the diode voltage by 130 mV. Thus VD = 560 mV.
10 V
4.30
If D2 has saturation current IS , then D1 , which is 10 times larger, has saturation current 10IS . Thus we can write
I R1
ID2 = IS eVD2 /VT ID1 = 10I S eVD1 /VT Taking the ratio of the two equations above, we have 7 1 (VD2 −VD1 )/VT ID2 1 V /V T = = e e = ID1 3 10 10 70 = 78.7 mV ⇒ V = 0.025 ln 3
D1
V1
D2
V2
To instead achieve V = 60 mV, we need
At 20◦ C:
I1 − I2 1 0.06/0.025 ID2 e = = = 1.1 ID1 I2 10
VR1 = V2 = 520 mV
Solving the above equation with I1 still at 10 mA, we find I2 = 4.76 mA. 4.28 We can write the following node equation at the diode anodes: ID2 = 10 mA − V /R ID1 = V /R We can write the following equation for the diode voltages: V = VD2 − VD1
R1 = 520 k 520 mV = 1 μA 520 k Since the reverse current doubles for every 10◦ C rise in temperature, at 40◦ C, I = 4 μA I=
ID2
40°C
4 µA
40 mV 1 µA 480
We can write the following diode equations:
520 mV
ID2 = IS eVD2 /VT
V2 = 480 + 2.3 × 1 × 25 log 4
ID1 = IS eVD1 /VT
= 514.6 mV
Taking the ratio of the two equations above, we have 10 mA − V /R ID2 = e(VD2 −VD1 )/VT = eV /V T = ID1 V /R To achieve V = 50 mV, we need 10 mA − 0.05/R ID2 = e0.05/0.025 = 7.39 = ID1 0.05/R
20°C
VR1 = 4 μA × 520 k = 2.08 V 1 μA 4 V2 = 560 − 2.3 × 1 × 25 log 4
At 0◦ C, I =
= 525.4 mV 1 VR1 = × 520 = 0.13 V 4
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V2
Chapter 4–9 1.0 V = VT ln = 57.6 mV 0.1 ⇒ VD = 757.6 mV
4.31 For a diode conducting a constant current, the diode voltage decreases by approximately 2 mV per increase of 1◦ C. A decrease in VD by 100 mV corresponds to a junction temperature increase of 50◦ C.
ID = 3 mA:
3 0.1
= 85 mV ⇒ VD = 785 mV
The power dissipation is given by
V = VT ln
PD = (10 A) (0.6 V) = 6 W
For the second diode, with
The thermal resistance is given by
ID = 1 A and VD = 700 mV, we have
50◦ C T = 8.33◦ C/W = PD 6W
4.32 Given two different voltage/current measurements for a diode, we can write
ID = 1.0 mA: 0.001 = −173 mV V = VT ln 1 ⇒ VD = 527 mV
ID1 = IS eVD1 /VT
ID = 3 mA:
ID2 = IS eVD2 /VT
0.003 1 ⇒ VD = 555 mV
= −145 mV
V = VT ln
Taking the ratio of the above two equations, we have ID1 = IS e(VD1 −VD2 )/VT ⇒ VD1 − VD2 ID2 ID1 = VT ln ID2 For ID = 1 mA, we have 1 × 10−3 A V = VT ln = −230 mV 10 A ⇒ VD = 570 mV For ID = 3 mA, we have 3 × 10−3 A V = VT ln = −202 mV 10 A ⇒ VD = 598 mV Assuming VD changes by –2 mV per 1◦ C increase in temperature, we have, for ±20◦ C changes: For ID = 1 mA, 530 mV ≤ VD ≤ 610 mV For ID = 3 mA, 558 mV ≤ VD ≤ 638 mV Thus the overall range of VD is between 530 mV and 638 mV.
For both ID = 1.0 mA and ID = 3 mA, the difference between the two diode voltages is approximately 230 mV. Since, for a fixed diode current, the diode voltage changes with temperature at a constant rate (–2 mV per ◦ C temp. increase), this voltage difference will be independent of temperature!
4.34
R 1 kΩ VDD
i
1V
v
IS = 10−15 A = 10−12 mA Calculate some points v = 0.6 V ,
i = IS ev /V T
= 10−12 e0.6/0.025 4.33 Given two different voltage/current measurements for a diode, we have
0.03 mA
ID1 = IS e(VD1 −VD2 )/VT ⇒ VD1 − VD2 ID2 ID1 = VT ln ID2
v = 0.65 V,
i 0.2 mA
v = 0.7 V,
i 1.45 mA
For the first diode, with ID = 0.1 mA and VD = 700 mV, we have ID = 1 mA:
Make a sketch showing these points and load line and determine the operating point. The points for the load line are obtained using ID =
VDD − VD R
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Chapter 4–10
i (mA)
4.35
R 1 k
2.0
Diode characteristic
1V
ID
VD
1.0 0.8 0.6 0.4 0.2 0
Load line IS = 10−15 A = 10−12 mA
0.4
0.6
0.8
1.0
v (V)
Use the iterative analysis procedure: 1 − 0.7 = 0.3 mA 1K ID 0.3 = 0.025 ln 2. VD = VT ln −12 IS 10 1. VD = 0.7 V, ID =
From this sketch one can see that the operating point must lie between v = 0.65 V to v = 0.7 V i For i = 0.3 mA, v = VT ln IS 3 = 0.025 × ln 10−12 = 0.661 V For i = 0.4 mA, v = 0.668 V Now we can refine the diagram to obtain a better estimate
0.4
i (mA)
= 0.6607 V
||
1 − 0.6607 = 0.3393 mA 1K 0.3393 = 0.6638 V 3. VD = 0.025 ln 10−12 ID =
1 − 0.6638 = 0.3362 mA 1K 0.3362 = 0.6635 V 4. VD = 0.025 ln 10−12 ID =
1 − 0.6635 = 0.3365 mA 1 k Stop here as we are getting almost same value of ID and VD ID =
0.35
4.36
Load line
500 1V
0.30 0.660
0.664
0.67
v (V)
From this graph we get the operating point i = 0.338 mA, v = 0.6635 V Now we compare graphical results with the exponential model. At i = 0.338 mA i 0.338 = 0.025 × ln v = VT ln IS 10−12 = 0.6637 V
i
v
1 − 0.7 = 0.6 mA 0.5 k b) Diode has 0.7 V drop at 1 mA current. Use Eq. (4.5): i2 v 2 = v 1 + 2.3VT log i1 a) ID =
1. v = 0.7 V 1 − 0.7 = 0.6 mA 1 i= 0.5 k
0.6 1
The difference between the exponential model and graphical results is = 0.6637 − 0.6635
2. v = 0.7 + 2.3 × 0.025 log
= 0.0002 V
= 0.6872 V 1 − 0.6872 = 0.6255 mA 2 i= 0.5
= 0.2 mV
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Chapter 4–11
0.6255 3. v = 0.7 + 2.3 × 0.025 log 1
N
= 0.6882 V
⫹
1 − 0.6882 = 0.6235 mA 3 i= 0.5 k 0.6235 4. v = 0.7 + 2.3 × 0.025 log 1 = 0.6882 V 4 i=
1 − 0.6882 = 0.6235 mA 0.5 k
Stop as we are getting the same result.
V
1 mA
⫺ The voltage drop across each pair of diodes is 1.3 V. ∴ Voltage drop across each diode 1.3 = = 0.65 V. Using 2 I2 = I1 e(V2 −V1 )/VT = 2e(0.65−0.7)/0.025
4.37 We first find the value of IS for the diode, given by IS = ID e−VD /VT with ID = 1 mA and VD = 0.75 V. This gives IS = 9.36 × 10−17 A. In order to have 3.3 V across the 4 series-connected diodes, each diode drop must be 0.825 V. Applying this voltage to the diode gives current ID = 20.1 mA. We can then find the resistor value using 15 V − 3.3 V R= = 582 20.1 mA 4.38 Constant voltage drop model: Using v D = 0.7 V, ⇒ iD1 =
V − 0.7 R
Using v D = 0.6 V, ⇒ iD2 =
V − 0.6 R
For the difference in currents to be only 1%, ⇒ iD2 = 1.01iD1 V − 0.6 = 1.01 (V − 0.7) V = 10.7 V
= 0.2707 mA Thus current through each branch is 0.2707 mA. The 1 mA will split in =
1 = 3.69 branches. 0.2707
Choose N = 4. There are 4 pairs of diodes in parallel. ∴ We need 8 diodes. Current through each pair of diodes 1 mA = 0.25 mA = 4 ∴ Voltage across each pair 0.25 = 2 0.7 + 0.025 ln 2 = 1.296 V SPECIAL NOTE: There is another possible design utilizing only 6 diodes and realizing a voltage of 1.313 V. It consists of the series connection of 4 parallel diodes and 2 parallel diodes.
4.40 Refer to Example 4.2.
For V = 3 V and R = 1 k:
(a)
3 − 0.7 = 2.3 mA = 1
At VD = 0.7 V,
iD1
At VD = 0.6 V,
iD2 =
3 − 0.6 = 2.4 mA 1
2.4 iD2 = = 1.04 iD1 2.3 Thus the percentage difference is 4%.
10 V 5 I 1.861 10 k 0.86 mA
10 0 1 mA 10 V0V 3
1 0.7 V
4.39 Available diodes have 0.7 V drop at 2 mA current since 2VD = 1.4 V is close to 1.3 V, use N parallel pairs of diodes to split the 1 mA current evenly, as shown in the figure next.
4
5 k
2 0.7 10 5 1.86 mA
10 V
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Chapter 4–12
(b)
(c)
10 V
3 V
5 k
ID2
I0
V 0.7 V
Cutoff
I V 10 k
ID2
10 k
3 V 10 V
V = 3 − 0.7 = 2.3 V
10 − (−10) − 0.7 = 1.29 mA 15 VD = −10 + 1.29 (10) + 0.7 = 3.6 V
ID2 =
I=
2.3 + 3 = 0.53 mA 10
(d)
3 V 4.41 (a)
I
Cutoff 3 V V 10 k V 3 V I
I =0A V = −3 V
3 V V = −3 + 0.7 = −2.3 V
4.42
3 + 2.3 10 = 0.53 mA
(a)
I=
Cutoff 1 V
(b)
D1 3 V
2 V
10 k
V
D2 2 k
V Cutoff
3 V
I V = 2 − 0.7
3 V I =0A V = 3 − I (10) = 3 V
= 1.3 V 1.3 − (−3) I= 2 = 2.15 mA
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I
Chapter 4–13
(b) See analysis on Fig. (b).
(b)
3 V
I = 0.133 mA V =0V
I
2 k
4.44
1 V
D1
2 V
10 10 5 k 0.7 V
V
V I
10 5 10 10 2.5 V
D2 Cutoff
20 k
V = 1 + 0.7 = 1.7 V
(a)
3 − 1.7 = 0.65 mA I= 2
5 k
4.43
V
5 k
0 mA
+3V ID2
12 k
(b)
I0
1.5 V
2.5 V
D1 ID2
Cutoff
(a) I =
2.5 − 0.7 = 0.072 mA 5 + 20
ON 0.7 V D2 V
V = 0.072 × 20 = 1.44 V
6 k
I =0
(b) The diode will be cut off, thus
V = 1.5 − 2.5 = −1 V
(a)
3 V 4.45
3V 30.7 0.383 mA 6
vI
R
6 k
0.7 V I 0.383 0.25 0.133 mA ON
iD
v I ,peak − 0.7 ≤ 40 mA R √ 120 2 − 0.7 R≥ = 4.23 k 40 √ Reverse voltage = 120 2 = 169.7 V. iD,peak =
D2
D1
ON 0(3) 12 0.25 mA
V0V
12 k 3 V
The design is essentially the same since the supply voltage 0.7 V
(b)
(a) ID2 =
3 − 0.7 − (−3) = 0.294 mA 12 + 6
V = −3 + 0.294 × 6 = −1.23 V Check that D1 is off: Voltage at the anode of D1 = V + VD2 = −1.23 + 0.7 = −0.53 V which keeps D1 off.
4.46 Use the exponential diode model to find the percentage change in the current. iD = IS ev /VT iD2 = e(V2 −V1 )/VT = ev /VT iD1 For +5 mV change we obtain
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Chapter 4–14
iD2 = e5/25 = 1.221 iD1
For a parallel combination of 10 diodes, equivalent resistance, Req , is
% change 1.221 − 1 iD2 − iD1 × 100 = × 100 = iD1 1 = 22.1% For –5 mV change we obtain iD2 = e−5/25 = 0.818 iD1 % change =
0.818 − 1 iD2 − iD1 × 100 × 100 = iD1 1
= −18.1%
2.7 = 0.27 10 If there is a single diode conducting all the 0.1 A current, the connection resistance needed for the single diode will be = 0.27 − 0.25 = 0.02 .
Req =
4.48 The dc current I flows through the diode VT giving rise to the diode resistance rd = and I the small-signal equivalent circuit is represented by
Rs
Maximum allowable voltage signal change when the current change is limited to ±10% is found as follows: The current varies from 0.9 to 1.1 iD2 = eV /VT iD1 For 0.9, V = 25 ln (0.9) = −2.63 mV For 1.1, V = 25 ln (1.1) = +2.38 mV For ±10% current change the voltage signal change is from –2.63 mV to +2.38 mV 4.47
0.1 A
rd
vs
vo
rd VT /I VT = vs = vs VT rd + RS VT + IRS + RS I 25 mV Now, v o = 10 mV × 25 mV + 103 I vo = vs ×
I
vo
1 mA 0.24 mV 0.1 mA 2.0 mV 1 μA 9.6 mV
For v o = Ten diode connected in parallel and fed with a total current of 0.1 A. So the current through each 0.1 diode = = 0.01 A 10 Small signal resistance of each diode
1 0.025 vs = vs × 2 0.025 + 103 I
⇒ I = 25 μA
4.49 As shown in Problem 4.48,
25 mV VT = 2.5 = = iD 0.01 A
VT 0.025 vo = = vi VT + RS I 0.025 + 104 I
Equivalent resistance, Req , of 10 diodes connected in parallel is given by
Here RS = 10 k
2.5 = 0.25 Req = 10 If there is one diode conducting 0.1 A current, then the small signal resistance of this diode
(1)
The current changes are limited ±10%. Using exponential model, we get iD2 = ev /VT = 0.9 to 1.1 iD1 iD2 and here v = 25 × 10−3 ln iD1
25 mV = 0.25 0.1 A This value is the same as of 10 diodes connected in parallel.
For 0.9, v = −2.63 mV
If 0.2 is the resistance for making connection, the resistance in each branch = rd + 0.2 = 2.5 + 0.2 = 2.7
The variation is –2.63 mV to 2.38 mV for ±10% current variation. Thus the largest symmetrical output signal allowed is 2.38 mV in amplitude. To
=
For 1.1, v = 2.38 mV
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Chapter 4–15
obtain the corresponding input signal, we divide this by (v o /v i ): vˆs =
2.38 mV v o /v i
(2)
Now for the given values of v o /v i calculate I and vˆ S using Equations (1) and (2) vo vi 0.5 0.1 0.01 0.001
I = 500 μA, I = 600 μA, I = 900 μA,
I in mA vˆ s in mV
I = 990 μA,
0.0025 0.0225 0.2475 2.4975
I = 1 mA
4.76 23.8 238 2380
vo vi vo vi vo vi vo vi vo vi
I = 100 μA,
= 1000 μA,
= 100 × 10−3 = 0.1 V/V = 500 × 10−3 = 0.5 V/V = 600 × 10−3 = 0.6 V/V = 900 × 10−3 = 0.9 V/V = 990 × 10−3 = 0.99 V/V
vo = 1000 × 10−3 = 1 V/V vi
4.51 4.50
I
1 mA C 2 vo D1
C1
D1
D2
D3
vi vo
I vi
R D2
When both D1 and D2 are conducting, the small-signal model is
D4
I
rd1 vi
vo rd2 a. The current through each diode is
where we have replaced the large capacitors C1 and C2 by short circuits: VT rd 2 I vo 1 m −I = = = VT VT vi rd 1 + rd 2 1m + I 1m−I Thus vo = I, vi
R R vo = = vi R + (2rd 2rd ) R + rd rd
where I is in mA vo =0 vi
vo = 1 × 10−3 = 0.001 V/V vi vo = 10 × 10−3 = 0.01 V/V I = 10 μA, vi I = 1 μA,
2VT VT 0.05 = = I I I 2 From the equivalent circuit rd =
I
Now I = 0 μA,
I : 2
0 μA ∞ 1 μA 50 k 10 μA 5 k 100 μA 500 1 mA 50 10 mA 5
vo vi 0 0.167 0.667 0.952 0.995 0.9995
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10 k
Chapter 4–16
rd
rd
From the results of (a) above, for I = 1 mA, v o /v i = 0.995; thus the maximum input signal will be
vo vI
rd
rd R
10 k
Equivalent Circuit b. For signal current to be limited to ±10% of I (I is the biasing current), the change in diode voltage can be obtained from the equation
vˆ i = vˆ o /0.995 = 1/0.995 = 1.005 V The same result can be obtained from the figure above where the signal across the two series diodes is 5 mV, thus vˆ i = vˆ o + 5 mV = 1 V + 5 mV = 1.005 V. See also the figure below.
iD = eVD /V T = 0.9 to 1.1 I v D = −2.63 mV to +2.32 mV ±2.5 mV so the signal voltage across each diode is limited to 2.5 mV when the diode current remains within 10% of the dc bias current. ∴ v o = 10 − 2.5 − 2.5 = 5 mV and i =
5 mV = 0.5 μA 10 k 2.5 mV
2.5 mV
4.52 i
I
vo 10 mV
R 2.5 mV
10 k
D1
2.5 mV
The current through each diode
vI
i2
0.5 μA = 0.25 μA = 2 The signal current i is 0.5 μA, and this is 10% of the dc biasing current.
D2
i1
i3
v1
v3
v 2
v4
D3 iO
vO
i4 R 10 k D4
I
∴ DC biasing current I = 0.5 × 10 = 5 μA c. Now I = 1 mA. ∴ ID = 0.5 mA Maximum current derivation 10%. 0.5 = 0.05 mA ∴ id = 10 and i = 2id = 0.1 mA.
I = 1 mA Each diode exhibits 0.7 V drop at 1 mA current. Using diode exponential model we have i2 v 2 − v 1 = VT ln i1
= 0.1 × 10
and v 1 = 0.7 V, i1 = 1 mA i ⇒ v = 0.7 + VT ln 1
=1V
= 700 + 25 ln(i)
∴ Maximum v o = i × 10 k
SEDRA-ISM: “P-CH04-031-053” — 2014/11/12 — 17:25 — PAGE 16 — #8
Chapter 4–17
Calculation for different values of v O : v O = 0, iO = 0, and the current I = 1 mA divide equally between the D3 , D4 side and the D1 , D2 side.
For I = 0.5 mA, the output will saturate at 0.5 mA ×10 k = 5 V. vo (V )
I = 0.5 mA 2
10
v = 700 + 25 ln(0.5) 683 mV
5
i1 = i2 = i3 = i4 =
v 1 = v 2 = v 3 = v 4 = 683 mV
5.68
10.7
v1 (V)
5
v I = −v 1 + v 3 + v O = −683 + 683 + 0 = 0 V For v O = 1 V, iO =
I 0.5 mA
10.7 5.68
From the circuit, we have
I 1 mA
1 = 0.1 mA 10 k
Because of symmetry of the circuit, we obtain i3 = i2 =
I iO + = 0.5 + 0.05 = 0.55 mA and 2 2
4.53 Representing the diode by the small-signal resistances, the circuit is
i4 = i1 = 0.45 mA
rd
i2 v 3 = v 2 = 700 + 25 ln = 685 mV 1 i4 = 680 mV v 4 = v 1 = 700 + 25 ln 1
vO (V)
iO i3 = i2 i4 = i1 v 3 = v 2 v 4 = v 1 (mA) (mA) (mA) (mV) (mV)
vi
v I = −v 1 + v 3 + v O (V)
0.5
0.5
683
683
0
+1
0.1
0.55
0.45
685
680
1.005
+2
0.2
0.6
0.4
∼ 687
677
2.010
+5
0.5
0.75
0.25
∼ 693
665
5.028
+9
0.9
0.95
0.05
∼ 699
∼ 625
9.074
0.995
0.005
∼ 700
568
10.09
9.99 0.999 0.9995 0.0005 ∼ 700
510
10.18
10
0
10.7
1
0
700
vo
VT rd ID
0
1
C
0
+ 9.9 0.99
v I = −v 1 + v 2 + v O = −0.680 +0.685 + 1 = 1.005 V Similarly, other values are calculated as shown in the table. The largest values of v O on positive and negative side are +10 V and −10 V, respectively. This restriction is imposed by the current I = 1 mA A similar table can be generated for the negative values. It is symmetrical. For v I > +10.7, v O will be saturated at +10 V and it is because I = 1 mA. Similarly, for v I < −10.7 V, v O will be saturated at −10 V.
Vo = Vi
1 sC 1 rd + sC
=
1 1 + sCrd
1 Vo = Vi 1 + jωCrd Phase shift = −tan−1
ωCrd 1
VT = −tan−1 ωC I For phase shift of −45◦ , we obtain −45 = −tan−1 2π × 100 × 103 × 10 0.025 × 10−9 × I ⇒ I = 157 μA Now I varies from
157 μA to 157 × 10 μA 10
Range is 15.7 μA to 1570 μA Range of phase shift is −84.3◦ to −5.71◦
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Chapter 4–18
4.55
4.54
V
V
R
IL
R
VO
VO
VO IL
(a)
=
VO rd = = + V R + rd
VT /I VT R+ I
Small-signal model
VT IR + VT
(a) From the small-signal model
V + − 0.7 V + − VO = . For no load, I = R R ∴
rd
R
VO = −I L (rd R) VO = − (rd R) IL
VT VO = V + VT + (V + − 0.7)
(b) At no load, ID =
R
rd =
V
rd
VO
VO mr d mVT = = mr d + R mVT + IR V + mVT = mVT + (V + − 0.7m) (c) For m = 1 VO VT = = 1.75 mV/V V + VT + V + − 0.7 For m = 4
=−
1 ID ID + V + − 0.7 VT
=−
VT × ID
=−
V + − 0.7 VT × VT + V + − 0.7 ID
For
1 VT +1 V + − 0.7
mV VO ≤5 IL mA
V + − 0.7 5 mV VT × ≤ ID VT + V + − 0.7 mA mV 15 − 0.7 25 ≤5 × ID mA 0.025 + 15 − 0.7 i.e.,
ID ≥ 4.99 mA ID 5 mA R=
VO mVT = = 8.13 mV/V V + mVT + 15 − m × 0.7
VT ID
VO 1 = − (rd R) = − 1 1 IL + rd R
Small-signal model
(b) If m diodes are in series, we obtain
V + − 0.7 R
15 − 0.7 V + − 0.7 = ID 5 mA
R = 2.86 k
SEDRA-ISM: “P-CH04-054-069” — 2014/11/12 — 15:21 — PAGE 18 — #1
Chapter 4–19
Diode should be a 5-mA unit; that is, it conducts 5 mA at VD = 0.7 V, thus 5 × 10−3 = IS e0.7/0.025 . ⇒ IS = 3.46 × 10−15 A (c) For m diodes connected in series we have ID =
So now
=−
=−
=−
I = ID + IL = 7.39 mA + 1 mA = 8.39 mA ∴R=
V + − 0.7m R
and rd =
IL = 1.5/1.5 = 1 mA
5 − 1.5 = 417 8.39 mA
Use a small-signal model to find voltage VO when the value of the load resistor, RL , changes:
VT ID
VO 1 = −(R mrd ) = − 1 1 IL + R mrd
1 ID ID + V + − 0.7m mVT
rd =
VT 0.025 = 3.4 = ID 7.39
When load is disconnected, all the current I flows through the diode. Thus ID = 1 mA VO = ID × 2rd
mV T ID
mV T +1 V + − 0.7m
= 1 × 2 × 3.4
V + − 0.7m mVT ID V + − 0.7m + mVT
= 6.8 mV With RL = 1 k, IL
4.56
1.5 V = 1.5 mA 1
IL = 0.5 mA
5 V
ID = −0.5 mA
I
VO = −0.5 × 2 × 3.4
R
= −3.4 mV
IL Vo
ID
With RL = 750 , IL
RL
1.5 = 2 mA 0.75
IL = 1 mA ID = −1 mA VO = −1 × 2 × 3.4 Diode has 0.7 V drop at 1 mA current VO = 1.5 V when RL = 1.5 k
= −6.8 mV With RL = 500 ,
ID = IS eV /V T
1.5 = 3 mA 0.5
1 × 10−3 = IS e0.7/0.025
IL
⇒ IS = 6.91 × 10−16 A
IL = 2.0 mA
Voltage drop across each diode =
1.5 = 0.75 V. 2
ID = −2.0 mA
∴ ID = IS eV /V T = 6.91 × 10−16 × e0.75/0.025
VO = −2 × 2 × 3.4
= 7.38 mA
= −13.6 mV
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Chapter 4–20
4.58
4.57
5 V I I
R 200
vO 1.5 V VO IL ID
D1 ID D2
IL RL
150 VO
IL varies from 2 to 7 mA
To supply a load current of 2 to 7 mA, the current I must be greater than 7 mA. So I can be only 10 mA or 15 mA. Now let us evaluate VO for both 10-mA and 15-mA sources. For the 10-mA source:
(a) Iteration #1: VD = 0.7 V VO = 2VD = 1.4 V IL =
VO 1.4 = 9.33 mA = RL 0.15
Since IL varies from 2 to 7 mA, the current ID will varry from 8 to 3 mA.
I=
Correspondingly, the voltage across each diode changes by VD where
Iteration #2:
3 = eVD /VT 8 3 = −24.5 mV ⇒ VD = 25 ln 8 and the output voltage changes by VO = 2 × VD = −49 mV With I = 15 mA, the diodes current changes from 13 to 8 mA. Correspondingly, the voltage across each diode changes by VD where 8 = eVD /VT 13
5 − 1.4 5 − VO = = 18 mA R 0.2 ID = I − IL = 18 − 9.33 = 8.67 mA 8.67 = 0.696 V VD = 0.7 + 0.025 ln 10
VO = 1.393 V IL = 9.287 mA 5 − 1.393 = 18.04 mA 0.2 ID = 18.04 − 9.287 = 8.753 mA I=
Iteration #3:
8.753 = 0.697 VD = 0.7 + 0.025 ln 10 VO = 1.393 V IL = 9.287
8 ⇒ VD = 25 ln = −12.1 mV 13
I = 18.04 mA
and the output voltage changes by
No further iterations are necessary and
VO = 2 × VD = −24.2 mV
VO = 1.39 V
which is less than half that obtained with the 10-mA supply. Thus, from the point of view of reducing the change in VO as IL changes, we choose the 15-mA supply. Note, however, that the price paid is increased power dissipation.
ID = 8.753
(b) With no load: Iteration #1: VD = 0.7 V VO = 1.4 V
SEDRA-ISM: “P-CH04-054-069” — 2014/11/12 — 15:21 — PAGE 20 — #3
Chapter 4–21 5 − 1.4 = 18 mA 0.2 ID = I = 18 mA
I=
Iteration #2:
18 = 0.715 V VD = 0.7 + 0.025 ln 10 VO = 1.429 V
ID = 17.18 mA Iteration #3: 17.18 = 0.714 V VD = 0.7 + 0.025 ln 10 VO = 1.428 V No further iterations are needed and
I = 17.85 mA ID = 17.85 mA
VO = 1.43 V
Iteration #3:
(e) From the above we see that as VSupply changes from 5 V to 3.232 V (a change of −35.4%) the output voltage changes from 1.39 V to 1.29 V (a change of −7.19%).
17.85 = 0.714 V VD = 0.7 + 0.025 ln 10 VO = 1.43 V I = 17.86 mA ID = 17.86 mA No further iterations are warranted and VO = 1.43 V (c) VO = 1.39 − 0.1 = 1.29 V 1.29 = 8.6 mA 0.15 1.29 = 0.645 V VD = 2 IL =
As VSupply changes from 5 V to 6.786 V (a change of +35.4%) the output voltage changes from 1.39 V to 1.43 V (a change of +2.88%). Thus the worst-case situation occurs when VSupply is reduced, and Percentage change in VO per 1% change in 7.19 = 0.2% VSupply = 35.4 4.59 VZ = VZ0 + IZT rz
ID = 10 × e(0.645−0.7)/0.025
(a) 10 = 9.6 + 0.05 × rz
= 1.11 mA
⇒ rz = 8
I = IL + ID = 8.6 + 1.11 = 9.71 mA
For IZ = 2IZT = 100 mA,
VSupply = VO + IR = 1.29 + 9.71 × 0.2
VZ = 9.6 + 0.1 × 8 = 10.4 V
= 3.232 V which is a reduction of 1.768 V or −35.4%. (d) For VSupply = 5 + 1.786 = 6.786 V,
P = 10.4 × 0.1 = 1.04 W (b) 9.1 = VZ0 + 0.01 × 30
Iteration #1:
⇒ VZ0 = 8.8 V
VD = 0.7 V
At IZ = 2IZT = 20 mA,
VO = 1.4 V
VZ = 8.8 + 0.02 × 30 = 9.4 V
IL = 9.33 mA
P = 9.4 × 20 = 188 mW
6.768 − 1.4 = 26.84 I= 0.2 ID = I − IL = 26.84 − 9.33 = 17.51 mA Iteration #2:
(c) 6.8 = 6.6 + IZT × 2 ⇒ IZT = 0.1 A
17.51 = 0.714 V VD = 0.7 + 0.025 ln 10
At IZ = 2IZT = 0.2 A,
VO = 1.428 V
P = 7 × 0.2 = 1.4 W
IL = 9.52 mA
(d) 18 = 17.6 + 0.005 × rz
I = 26.70 mA
VZ = 6.6 + 0.2 × 2 = 7 V
⇒ rz = 80
SEDRA-ISM: “P-CH04-054-069” — 2014/11/12 — 15:21 — PAGE 21 — #4
Chapter 4–22
At IZ = 2IZT = 0.01 A,
4.63
VZ = 17.6 + 0.01 × 80 = 18.4 V
10 V
P = 18.4 × 0.01 = 0.184 W = 184 mW (e) 7.5 = VZ0 + 0.2 × 1.5
I
⇒ VZ0 = 7.2 V
R
At IZ = 2IZT = 0.4 A,
VO VZ
VZ = 7.2 + 0.4 × 1.5 = 7.8 V P = 7.8 × 0.4 = 3.12 W
IZ
IL
RL
4.60 (a) Three 6.8-V zeners provide 3 × 6.8 = 20.4 V with 3 ×10 = 30- resistance. Neglecting R, we have Load regulation = −30 mV/mA.
(a)
(b) For 5.1-V zeners we use 4 diodes to provide 20.4 V with 4 ×30 = 120- resistance. Load regulation = −120 mV/mA
11 V
4.61
82
I
R
vS
VO VZ
8
vO rZ
Small-signal model
VZ0
From the small-signal model we obtain
(b)
8 8 v O = = v S 8 + 82 90
9 V
Now v S = 1.0 V. 8 8 v S = × 1.0 ∴ v O = 90 90 = 88.9 mV
I
R
4.62 VZ = VZ0 + IZT rZ
VO VZ
9.1 = VZ0 + 0.02 × 10 ⇒ VZ0 = 8.9 V At IZ = 10 mA,
0.5 mA
IL
RL
VZ = 8.9 + 0.01 × 10 = 9.0 V At IZ = 50 mA, VZ = 8.9 + 0.05 × 10 = 9.4 V
(c)
SEDRA-ISM: “P-CH04-054-069” — 2014/11/12 — 15:21 — PAGE 22 — #5
Chapter 4–23
To obtain VO = 7.5 V, we must arrange for IZ = 10 mA (the current at which the zener is specified).
4.64
9V±1V
Now,
R
VO 7.5 = = 5 mA RL 1.5
IL = Thus
VO
I = IZ + IL = 10 + 5 = 15 mA and
IZ
10 − 7.5 10 − VO = = 167 R= I 15 When the supply undergoes a change VS , the change in the output voltage, VO , can be determined from VO (RL rz ) = VS (RL rz ) + R 1.5 0.03 = 0.15 = (1.5 0.03) + 0.167
GIVEN PARAMETERS VZ = 6.8V, rz = 5 IZ = 20 mA
For VS = +1 V (10% high), VO = +0.15 V and VO = 7.65 V.
At knee,
For VS = −1 V (10% low), VO = −0.15 V and VO = 7.35 V.
rz = 750
When the load is removed and VS = 11 V, we can use the zener model to determine VO . Refer to Fig. (b). To determine VZ0 , we use VZ = VZ0 + IZT rz 7.5 = VZ0 + 0.01 × 30 ⇒ VZ0 = 7.2 V From Fig. (b) we have I=
11 − 7.2 = 19.3 mA 0.167 + 0.03
Thus
IZK = 0.25 mA
FIRST DESIGN: 9-V supply can easily supply current Let IZ = 20 mA, well above knee. ∴ R=
9 − 6.8 = 110 20
Line regulation = =
rZ VO = VS rZ + R
5 5 + 110
= 43.5
mV V
VO = VZ0 + Irz
SECOND DESIGN: limited current from 9-V supply
= 7.2 + 0.0193 × 30 = 7.78 V
IZ = 0.25 mA
To determine the smallest allowable value of RL while VS = 9 V, refer to Fig. (c). Note that IZ = 0.5 mA, thus
VZ = VZK VZO − calculate VZ0 from VZ = VZ0 + rZ IZT
VZ = VZK VZ0 = 7.2 V
6.8 = VZ0 + 5 × 0.02
I=
9 − 7.2 = 10.69 mA 0.167
IL = I − IZ = 10.69 − 0.5 = 10.19 mA RL =
VO 7.2 = = 707 IL 10.19
VO = 7.2 V
VZ0 = 6.7 V ∴R=
8 − 6.7 = 5.2 k 0.25
LINE REGULATION = = 126
750 VO = VS 750 + 5200
mV V
SEDRA-ISM: “P-CH04-054-069” — 2014/11/12 — 15:21 — PAGE 23 — #6
Chapter 4–24
If IL is reduced by 50%, then
4.65
VS 15 V 10%
I
I=
R
IL
15 − VO 0.3 VO − 8.74 0.04
IZ =
VO VZ
IZ
1 9.15 × = 4.6 mA 2 1
IL =
15 − VO VO − 8.74 = + 4.6 0.3 0.04
RL
⇒ VO = 9.31 V which is an increase of 0.16 V. When the supply voltage is low, VS = 13.5 V
VZ = VZ0 + IZT rz
⇒ VZ0 = 8.74 V
and RL is at its lowest value, to maintain regulation, the zener current must be at least equal to IZK , thus
For IZ = 10 mA,
IZ = 0.5 mA
VZ = 8.74 + 0.01 × 40 = 9.14 V
VZ = VZK VZ0 8.74
9.14 = 9.14 mA IL = 1 k
I=
9.1 = VZ0 + 0.009 × 40
I = IZ + IL = 10 + 9.14 = 19.14 mA
IL = I − IZ = 15.87 − 0.5 = 15.37 mA
15 − 9.14 = 306 R= 19.4
RL =
Select R = 300
15 − VO 0.3
Line regulation = (1)
IL =
VO 1
(2)
IZ =
VO − VZ0 VO − 8.74 = rz 0.04
(3)
Since I = IZ + IL , we can use (1)–(3) to obtain VO : VO − 8.74 15 − VO = + VO 0.3 0.04 ⇒ VO = 9.15 V VO = VS = ±1.5 ×
rz RL (rz RL ) + R
(0.04 1) (0.04 1) + 0.3
= ±0.17 V
VZ 8.74 = 589 = IL 15.37
The lowest value of output voltage = 8.74 V
Denoting the resulting output voltage VO , we obtain I=
13.5 − 8.74 = 15.87 mA 0.3
170 mV 1.5 V
= 113 mV/V Load regulation = −(rz R) = −(40 300) = −35 mV/mA Or using the results obtained in this problem: For a reduction in IL of 4.6 mA, VO = +0.16 V, thus Load regulation = −
160 = −35 mV/mA 4.6
4.66 (a) VZT = VZ0 + rz IZT 10 = VZ0 + 7 (0.025) ⇒ VZ0 = 9.825 V
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Chapter 4–25
(b) The minimum zener current of 5 mA occurs when IL = 20 mA and VS is at its minimum of 20(1 − 0.25) = 15 V. See the circuit below:
25 V
VS 15 V
207
VZ
R IL 0 20 mA VO IZmin 5 mA
IZ RL 25 − VZ 207 207VZ = 207 (9.825) + 7 (25) − 7VZ = 9.825 + 7 ×
R≤
15 − VZ0 20 + 5
⇒ VZ = 10.32 V
where we have used the minimum value of VS , the maximum value of load current, and the minimum required value of zener diode current, and we assumed that at this current VZ VZ0 . Thus, 15 − 9.825 + 7 R≤ 25
25 − 10.32 = 70.9 mA 0.207 PZ = 10.32 × 70.9 IZmax =
= 732 mW
4.67
≤ 207 . ∴ use R = 207 7 mV (c) Line regulation = = 33 207 + 7 V ±25% change in v S ≡ ± 5 V
vS
vO
R
VO changes by ±5 × 33 = ±0.165 mV ±0.165 × 100 = ±1.65% corresponding to 10 (d) Load regulation = − (rZ R)
Using the constant voltage drop model:
= −(7 207) = −6.77
ideal
VD 0.7 V
or –6.77 V/A
VO = −6.77 × 20 mA = −135.4 mV corresponding to −
0.1354 × 100 = −1.35% 10
R 1 k
vS
(e) The maximum zener current occurs at no load IL = 0 and the supply at its largest value of 20 +
1 (20) = 25 V. 4
VZ = VZ0 + rZ IZ
vO
(a) v O = v S + 0.7 V, v O = 0,
For v S ≤ − 0.7 V
for v S ≥ −0.7 V
SEDRA-ISM: “P-CH04-054-069” — 2014/11/12 — 15:21 — PAGE 25 — #8
Chapter 4–26
iD = IS ev D /V T
vO 0.7 V 0
iD = e[v D −v D (at 1 mA)]/V T iD (1 mA) iD v D − v D (at 1 mA) = VT ln 1 mA v O /R v D = v D (at 1 mA) + VT ln 1
vS
slope 1
vO = vS − vD (b)
= v S − v D (at 1 mA) − VT ln
O
R
where R is in k.
vS
10 V
v
vO 4.69
t
0.7 V
0.7 V
10 V
0.7 V vS
R 1 k vO
u
(c) The diode conducts at an angle 0.7 = 4◦ and stops θ = sin−1 10
vS (V) 2.5 1.8
at π − θ = 176◦ Thus the conduction angle is π − 2θ
0.7 0
= 172◦ or 3 rad. v O,avg =
−1 2π
π −θ
t1
(10 sin φ − 0.7) d φ
T 4
t2 T 2
T
θ
2.5
−1 [−10 cos φ − 0.7φ]πθ −θ = 2π = −2.85 V
First find t1 and t2
(d) Peak current in diode is
0.7 2.5 = T t1 4 ⇒ t1 = 0.07 T
10 − 0.7 = 9.3 mA 1 (e) PIV occurs when v S is at its the peak and v O = 0.
t2 =
PIV= 10 V
T − t1 2
T − 0.07 T 2 t2 = 0.43 T =
4.68 D
vD vS
R
vO
1 × area of shaded triangle T T × (2.5 − 0.7) × − t1 4 1 × 1.8 × T − 0.07 4
v O (ave.) = =
1 T
=
1 T
= 0.324 V
SEDRA-ISM: “P-CH04-054-069” — 2014/11/12 — 15:21 — PAGE 26 — #9
t
Chapter 4–27 √ vˆO = 10 2 − VD = 13.44 V
4.70
Conduction starts at θ = sin−1
ideal 0.7 V
12 : 1
vS
10 Vrms
1 k vO
120 Vrms 60 Hz √ vˆO = 10 2 − 0.7 = 13.44 V
2.84◦ = 0.05 rad and ends at π − θ. Conduction angle = π − 2θ = 3.04 rad in each half cycle. Thus the fraction of a cycle for which one of the two 2(3.04) diodes conduct = × 100 = 96.8% 2π Note that during 96.8% of the cycle there will be conduction. However, each of the two diodes conducts for only half the time, i.e., for 48.4% of the cycle. v O,avg =
Conduction begins at √ 10 2 sin θ = 0.7 0.7 θ = sin−1 = 2.84◦ √ 10 2
1 π
iL,avg =
Conduction ends at π − θ .
4.72
12 : 1 D4
The diode conducts for 3.04 × 100 = 48.4% of the cycle 2π 1 2π
θ
8.3 = 8.3 mA 1 k
∴ Conduction angle = π − 2θ = 3.04 rad
v O,avg =
π−θ √ (10 2sinφ − 0.7)d φ
= 8.3 V
= 0.0495 rad
π −θ
0.7 √ = 10 2
10 Vrms vs
120 Vrms
√ (10 2sinφ − 0.7) d φ
D1 R 1 k D3
D2
VD 0.7 V
θ
√ Peak voltage across R = 10 2 − 2VD √ = 10 2 − 1.4
= 4.15 V v O,avg iD,avg = = 4.15 mA R
= 12.74 V 4.71
vS D1
6:1
10 2 V 1 k vo
10 Vrms 120 Vrms 60 Hz
1.4 V t
10 Vrms D2 θ = sin−1
1.4 √ = 5.68◦ = 0.1 rad 10 2
Fraction of cycle that D1 & D2 conduct is
vS , vO (V) vs vo
10 V 0.7 V 0
t
π − 2θ × 100 = 46.8% 2π Note that D3 & D4 conduct in the other half cycle so that there is 2 (46.8) = 93.6% conduction interval. v O,avg =
2 2π
π−θ
√ (10 2sinφ − 2V D ) d φ
θ
SEDRA-ISM: “P-CH04-070-085” — 2014/11/12 — 11:17 — PAGE 27 — #1
Chapter 4–28
=
√ √ 4.75 120 2 ± 10%: 20 2 ± 10%
4.73
⇒ Turns ratio = 6:1 √ 20 2 vS = ± 10% 2 PIV= 2Vs − VD √ 20 2 × 1.1 − 0.7 =2× 2 = 30.4 V
π −θ √ 1 −12 2 cos φ − 1.4φ θ π √ 2(12 2 cos θ ) 1.4 (π − 2θ) − = π π = 7.65 V v O,avg 7.65 = = 7.65 mA iR,avg = R 1
vS vS
120 Vrms
R
vO
Using a factor of 1.5 for safety, we select a diode having a PIV rating of approximately 45 V.
4.76 The circuit is a full-wave rectifier with center tapped secondary winding. The circuit can − be analyzed by looking at v + O and v O separately.
Refer to Fig. 4.24.
vS vS
For VD Vs , conduction angle π, and 2 2 v O,avg = Vs − VD = Vs − 0.7 π π (a) For v O,avg = 10 V π × 10.7 = 16.8 V 2 √ 120 2 Turns ratio = = 10.1 to 1 16.8 (b) For v O,avg = 100 V
D1
D3
Vs =
vO
vS vS
π × 100.7 = 158.2 V 2 √ 120 2 Turns ratio = = 1.072 to 1 158.2 Vs =
D4
R
D2
4.74 Refer to Fig. 4.25 For 2VD Vs 2 2 Vs − 2VD = Vs − 1.4 π π (a) For VO,avg = 10 V VO,avg =
2 · Vs − 1.4 π π ∴ Vˆ s = 11.4 = 17.9 V 2 √ 120 2 Turns ratio = = 9.5 to 1 17.9 (b) For VO,avg = 100 V 10 V =
2 · Vs − 1.4 π π = 159 V ⇒ Vs = 101.4 2 √ 120 2 Turns ratio = = 1.07 to 1 159
100 V =
v O, avg =
1 2π
(VS sinφ − 0.7) d φ = 12
2Vs − 0.7 = 12 π where we have assumed Vs 0.7 V and thus the conduction angle (in each half cycle) is almost π . =
Vs =
12 + 0.7 π = 19.95 V 2
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Chapter 4–29
= 2VS − 0.7
(b) (i) Using Eq (4.30), we have the conduction angle =
ωt ∼ = 2Vr / Vp − VD
2 × 0.1 Vp − 0.7
= Vp − 0.7 √ = 0.2
= 39.2 V
= 0.447 rad
If choosing a diode, allow a safety margin by moving a factor of 1.5, thus
∴ Fraction of cycle for
Thus voltage across secondary winding = 2VS 40 V Looking at D4 , PIV= VS − VO− = VS + (VS − 0.7)
conduction =
PIV 60 V
= 7.1% 4.77
0.447 × 100 2π
(ii) ωt 2 × 0.01 R
12 : 1
120 Vrms
10 Vrms VS
R 1 k
C vO
Vp Vr
VpVD
Vp − 0.7 = 0.141 rad Vp − 0.7
0.141 × 100 = 2.24% 2π (c) (i) Use Eq (4.31): ⎛
⎞ 2 Vp − VD ⎠ iD,avg = IL ⎝1 + π Vr Fraction of cycle =
2 Vp − VD v O,avg
1+π = R 0.1 Vp − VD 2 12.77 1 + π = 0.1 103 = 192 mA
√ 13.37 1 + π 200 3 10
T (i) Vr ∼ [Eq. (4.28)] = Vp − VD CR
T 0.1 Vp − VD = Vp − VD CR 1 = 166.7 μF C= 0.1 × 60 × 103
(ii) iD,avg =
(ii) For
2 12.77 1 + 2π = 0.1 103
Vp − VD T Vr = 0.01 Vp − VD = CR C = 1667 μF 1 (a) (i) v O, avg = Vp − VD − VΓ 2 √ √ 1 = 10 2 − 0.7 − 10 2 − 0.7 0.1 2 √ 0.1 = 10 2 − 0.7 1 − 2 = 12.77 V
√ 0.01 (ii) v O, avg = 10 2 − 0.7 1 − 2 = 13.37 V
= 607 mA (d) Adapting ⎛ Eq. (4.32), we obtain ⎞ 2 Vp − VD ⎠ (i) iD,peak = IL ⎝1 + 2π Vr
= 371 mA (ii) iD,peak
13.37 2 = 1 + 2π 0.01 103
= 1201 mA 1.2 A
Vp − VD 4.78 (i) Vr = 0.1 Vp − VD = 2fCR The factor of 2 accounts for discharge occurring 1 . only during half of the period, T /2 = 2f
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Chapter 4–30
C=
1 1 = 83.3 μF = (2fR) 0.1 2 (60) 103 × 0.1
(ii) C =
1 = 833 μF 2 (60) × 103 × 0.01
1 (a) (i) VO = Vp − VD − Vr 2
0.1 = Vp − VD 1 − 2 0.1 = (13.44) 1 − 2 = 12.77 V
0.01 (ii) VO = (13.44) 1 − 2 (b) (i) Fraction of cycle = =
2Vr / Vp − VD π
= 13.37 V
2ωt × 100 2π
Vp − VD 2Vr
12.77 1 = 1+π = 102.5 mA 1 2 (0.1) 1 13.37 1+π√ (ii) iD, avg = 1 2 (0.01) = 310 mA (d) Use Eq. (4.35): 1 = 192 mA (i) ˆiD = IL 1 + 2π √ 2 (0.1) 1 = 607 mA (ii) ˆiD = IL 1 + 2π √ 0.02
Vp − 2VD 4.79 (i) Vr = 0.1 Vp − VD × 2 = 2fCR
Vp − 2VD 1
= 83.3 μF C= Vp − 2VD 2 (0.1) fR (ii) C =
1 = 833 μF 2 (0.01) fR
1 (a) VO = Vp − 2VD − Vr 2
2ωt × 100 2π
√ 2 (0.1) × 100 = 14.2% π √ 2 (0.01) (ii) Fraction of cycle = × 100 = 4.5% π 12.1 1 1+π = 97 mA (c) (i) iD, avg = 1 0.2
√ 12.68 1 + π/ 0.02 = 249 mA 1 12.1 1 1 + 2π = 182 mA (d) (i) ˆiD = 1 0.2 (ii) iD, avg =
1 2 (0.1) × 100 = 14.2% π √ 2 2 (0.01) (ii) Fraction of cycle = × 100 2π = 4.5%
(i) iD, avg = IL 1 + π
= Vp − 2VD × 0.95 √ = (10 2 − 2 × 0.7) × 0.95 = 12.1 V √ (ii) VO = (10 2 − 2 × 0.7) × 0.995 = 12.68 V
=
=
1 Vp − 2VD × 0.1 2
(b) (i) Fraction of cycle =
× 100
(c) Use Eq. (4.34):
(i) VO = Vp − 2VD −
12.68 1 1 + 2π = 576 mA (ii) ˆiD = 1 0.02
4.80
0.7 V
120 Vrms 60 Hz
vS
R 200
VO = 12 V ± 1 V (ripple) RL = 200 (a) VO = Vp − VD − 1 ⇒ Vp = 13 + 0.7 = 13.7 V 13.7 Vrms = √ = 9.7 V 2 (b) Vr = 2=
Vp − VD fCR
13.7 − 0.7 60 × C × 200
⇒C=
13 = 542 μF 2 × 60 × 200
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C vO
Chapter 4–31
This voltage appears across each half of the transformer secondary. Across the entire secondary we have 2 × 9.7 = 19.4 V (rms).
Vp VpVD
t
Vp
PIV
(c) When the diode is cut off, the maximum reverse voltage across it will occur when v S = −Vp . At this time, v O = VO and the maximum reverse voltage will be
PIV
Maximum reverse voltage = VO + Vp
(b) Vr =
= 12 + 13.7 = 25.7 V
2=
Using a factor of safety of 1.5 we obtain PIV = 1.5 × 25.7 = 38.5 V
(d) iDav = IL 1 + π
2(Vp − VD ) Vr
In specifying the PIV for the diodes, one usually uses a factor of safety of about 1.5,
2(Vp − VD ) (e) iDmax = IL 1 + 2π Vr 2(13.7 − 0.7) 12 1 + 2π = 0.2 2
PIV = 1.5 × 25.7 = 38.5 V Vp − VD (d) iDav = IL 1 + π 2 Vr 13.7 − 0.7 12 1+π = 0.2 2×2
= 399 mA
Vp − VD 2 Vr 13.7 − 0.7 12 1 + 2π = 0.2 2×2
4.81
D1
Vs 0.7 V Vs 0.7 V
= 739 mA
C
R
D2 (a) VO = Vp − VD − 1 ⇒ Vp = VO + VD + 1 = 13 + 0.7 = 13.7 V Vrms
(e) iDmax = IL 1 + 2π
= 1.42 A
120 Vrms 60 Hz
12 = 271 μF 2 × 2 × 60 × 200
(c) Maximum reverse voltage across D1 occurs when v S = −Vp . At this point v O = VO . Thus maximum reverse voltage = VO + Vp = 12 + 13.7 = 25.7. The same applies to D2 .
2(Vp − VD ) VO 1+π = RL Vr 2(13.7 − 0.7) 12 1+π = 0.2 2 = 739 mA
13.7 − 0.7 2 × 60 × 200 × C
⇒C=
Vp − VD 2fCR
13.7 = √ = 9.7 V 2
vO
4.82
120 Vrms 60 Hz
D4
D1
C R
vS
vO D2
D3
(a) VO = Vp − 2VD − 1 ⇒ Vp = VO +2VD +1 = 12+2×0.7+1 = 14.4 V
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Chapter 4–32
12 V 11.3 11.3 Vr 10.2 V
14.4 Vrms = √ = 10.2 V 2 Vp − 2 VD (b) Vr = 2fCR 14.4 − 1.4 = 271 μF 2 × 2 × 60 × 200
⇒C=
During the diode’s off interval (which is almost equal to T ) the capacitor discharges and its voltage is given by v O (t) = 11.3 e−t/CR where C = 100 μF and R = 100 , thus
PIV = 1.5 × 13.7 = 20.5 V Vp − 2 VD (d) iDav = IL 1 + π 2 Vr 14.4 − 1.4 12 1+π = 0.2 2×2
Vp − 2 VD (e) iDmax = IL 1 + 2π 2 Vr 14.4 − 0.7 12 1 + 2π = 0.2 2×2
4
(c)
The same applies to the other three diodes. In specifying the PIV rating for the diode we use a factor of safety of 1.5 to obtain
Vr 1.13 V
t
Maximum reverse voltage = −Vp + VD3 = −14.4 + 0.7 = −13.7 V
T
0
(c) The maximum reverse voltage across D1 occurs when Vs = −Vp = −14.4 V. At this time D3 is conducting, thus
= 400 mA
}
CR = 100 × 10−6 × 100 = 0.01 s At the end of the discharge interval, t T and v O = 11.3 e−T /CR Since T = 0.001 s is much smaller than CR, T v O 11.3 1 − CR
The ripple voltage Vr can be found as T Vr = 11.3 − 11.3 1 − CR 11.3 × 0.001 11.3T = = 1.13 V CR 0.01 The average dc output voltage is =
= 740 mA
Vr 1.13 = 11.3 − = 10.74 V 2 2 To obtain the interval during which the diode conducts, t, refer to Fig. (c). v O = 11.3 −
4.83
vI
C 100 F
R 100
vO
Vr 12 = T /4 t
(a)
⇒ t =
1.13 × 1 Vr × (T /4) = 12 12 × 4
= 23.5 μs
vI 12 V 11.3 V
Vr
vO
Now, using the fact that the charge gained by the capacitor when the diode is conducting is equal to the charge lost by the capacitor during its discharge interval, we can write
vI
t
iCav × t = C Vr ⇒ iCav =
12 V
C Vr 100 × 10−6 × 1.13 = 4.8 A = t 23.5 × 10−6
T
(b)
1 ms
iDav = iCav + iLav where iLav is the average current through R during the short interval t. This is approximately 11.3 11.3 = = 0.113 A. Thus R 100
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Chapter 4–33
Finally, to obtain the peak diode current, we use
iDmax = IL (1 + 2π (Vp − 0.7)/2 Vr ) = 0.1(1 + 2π 12.5/2 )
iDmax = iCmax + iLmax
= 1.671 A
dv I 11.3 =C + dt R 11.3 12 + =C× T /4 R
To determine the required PIV rating of each diode, we determine the maximum reverse voltage that appears across one of the diodes, say D1 . This occurs when v S is at its maximum negative value −Vp . Since the cathode of D1 will be at +12.5 V, the maximum reverse voltage across D1 will be 12.5 + 13.2 = 25.7 V. Using a factor of safety of 1.5, then each of the four diodes must have
iDav = 4.8 + 0.113 = 4.913 A
= 100 × 10−6 ×
12 × 4 11.3 + 1 × 10−3 100
= 4.8 + 0.113 = 4.913 A which is equal to the average value. This is a result of the linear v I which gives rise to a constant capacitor current during the diode conduction interval. Thus iCmax = iCav = 4.8 A. Also, the maximum value of iL is approximately equal to its average value during the short interval t. 4.84 Refer to Fig. P4.76 and let a capacitor C be connected across each of the load resistors R. The − two supplies v + O and v O are identical. Each is a full-wave rectifier similar to that of the tapped-transformer circuit. For each supply, VO = 12 V Vr = 1 V (peak to peak) Thus v O = 12 ± 0.5 V It follows that the peak value of v S must be 12.5 + 0.7 = 13.2 V and the total rms voltage across the secondary will be =
2 × 13.2 = 18.7 V (rms) √ 2
120 = 6.43:1 18.7 To deliver 100-mA dc current to each load,
Transformer turns ratio =
12 = 120 0.1 Now, the value of C can be found from
PIV = 1.5 × 25.7 = 38.6 V 4.85 Refer to Fig. P4.85. When v I is positive, v A goes positive, turning on the diode and closing the negative feedback loop around the op amp. The result is that v − = v I , v O = 2v − = 2v I , and v A = v O + 0.7. Thus (a) v I = +1 V, v − = +1 V, v O = +2 V, and v A = +2.7 V. (b) v I = +3 V, v − = +3 V, v O = +6 V, and v A = +6.7 V. When v I goes negative, v A follows, the diode turns off, and the feedback loop is opened. The op amp saturates with v A = −13 V, v − = 0 V and v O = 0 V. Thus (c) v I = −1 V, v − = 0 V, v O = 0 V, and v A = −13 V. (d) v I = −3 V, v − = 0 V, v O = 0 V, and v A = −13 V. Finally, if v I is a symmetrical square wave of 1-kHz frequency, 5-V amplitude, and zero average, the output will be zero during the negative half cycles of the input and will equal twice the input during the positive half cycles. See figure.
vO
R=
Vr = 1=
Vp − 0.7 2fCR
5 V
12.5 2 × 60 × C × 120
⇒ C = 868 μF To specify the diodes, we determine iDav and iDmax , iDav = IL (1 + π (Vp − 0.7)/2 Vr ) = 0.1(1 + π 12.5/2 ) = 785 mA
10 V
0
vI
t
5 V 1 ms
Thus, v O is a square wave with 0-V and +10-V levels, i.e. 5-V average and, of course, the same frequency (1 kHz) as the input.
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Chapter 4–34
(b) See figure (b) on next page. Here v O = v I for v I ≥ 2.5 V. At v I = 2.5 V, v O = 2.5 V and the diode begins to conduct. The diode will be conducting 1 mA and exhibiting a drop of 0.7 at v O = 2.3 V. The corresponding value of v I
4.86 v I > 0: D1 conducts and D2 cutoff v I < 0: D1 cutoff, vO = −1 D2 conducts ∼ vI
v I = v O − iR = 2.3 − 1 × 1 = +1.3 V
vO
slope 1
As v I decreases below 1.3 V, the diode current increases, but the diode voltage remains constant at 0.7 V. Thus v O flattens at about 2.3 V.
v A = −0.7 V
(c) See figure (c) on next page. For v I ≤ −2.5 V, the diode is off, and v O = v I . At v I = −2.5 V the diode begins to conduct and its current reaches 1 mA at v I = −1.3 V (corresponding to v O = −2.3 V). As v I further increases, the diode current increases but its voltage remains constant at 0.7 V. Thus v O flattens, as shown.
Keeps D2 off so no current flows through R
(d) See figure (d) on next page.
vI
(a) v I = +1 V vO = 0 V
⇒ v− = 0 V Virtual ground as feedback loop is closed through D1
4.88
3 V
(b) v I = +3 V vO = 0 V
D1
v A = −0.7 V
R = 0.5 k
v− = 0 V
vI
(c) v I = −1 V
vo i
v O = +1 V
D2
v A = 1.7 V v− = 0 V ∼ Virtual ground as negative feedback loop is closed through D2 and R.
3 V (a)
(d) v I = −3 V ⇒ v O = +3 V v A = +3.7 V v− = 0 V
4.87 (a) See figure (a) on next page. For v I ≤ 3.5 V, i = 0 and v O = v I . At v I = 3.5 V, the diode begins to conduct. At v O = 3.7 V, the diode is conducting i = 1 mA and thus v I = v O + i × 1 k = 4.7 V For v I > 4.7 V the diode current increases but the diode voltage remains constant at 0.7 V, thus v O flattens and v O vs. v I becomes a horizontal line. In practice, the diode voltage increases slowly and the line will have a small nonzero slope.
From Fig. (a) we see that for −3.5 V ≤ v I ≤ +3.5 V, diodes D1 and D2 will be cut off and i = 0. Thus, v O = v I . For v I ≥ +3.5 V, diode D1 begins to conduct and its voltage reaches 0.7 V (and thus v O = +3.7 V) at i = 1 mA. The corresponding value of v I is v I = v O − iR v I = 3.7 + 1 × 0.5 = +4.2 V For v I ≥ 4.2 V, the voltage of diode D1 remains 0.7 V and v O saturates at +3.7 V. A similar description applies for v I ≤ −3.5 V. Here D2 conducts at v I = −3.5 V and its voltage becomes 0.7 V, and hence v O = −3.7 V, at i = 1 mA (in the direction into v I ) at v I = −4.2 V. For v I ≤ −4.2 V, v O = −3.7 V.
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Chapter 4–35
These figures belong to Problem 4.87.
vO (V)
3.7 3.5
3 V
D i vI (V)
vO
vI R 1 k
slope 1
3.5
4.7
(a) vO (V)
3 V
slope 1
2.5 2.3 D
i
vO
vI R 1 k
1.3
vI (V)
2.5
(b) vO (V) 2.5 vI
R 1 k
1.3
vI (V)
vO
i
2.3
D
2.5 3 V slope 1 (c) vO (V)
vI
R 1 k i
vO
4.7
3.5
vI (V)
D slope 1 3.5
3 V
3.7 (d)
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Chapter 4–36
This figure belongs to Problem 4.88, part b.
vO (V) 3.7 3.5
4.2
3.5 3.5
0
4.2
vI (V)
slope 1 (Not to scale)
3.5 3.7
D1 and D2 OFF
D1 OFF D2 ON
D1 ON D2 OFF
(b)
Figure (b) shows a sketch of the transfer characteristic of this double limiter.
4.90
1 k
vI
vO
4.89 See figure.
D4
3 V
D1 Z
D1
D2
vI
vO R 0.5 k
D2
D3
(a) (a)
vO (V) 3.7 3.5
slope 1
2.5 2.3
(Not to scale)
The limiter thresholds and the output saturation levels are found as 2 × 0.7 + 6.8 = 8.2 V. The transfer characteristic is given in Fig. (b). See figure on next page.
1 k
4.91 1.8 2.5 D1 ON D2 OFF
3.5 4.2
D1 & D2 OFF
vI (V)
vI
vO D1
D2
D1 OFF D2 ON
(b)
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Chapter 4–37
This figure belong to Problem 4.90, part b.
Z
Z
All diodes and zener are OFF
vO (V)
Diodes have 0.7 V drop at 1 mA current ∴ For diode D1 iD = e(v O −0.7)/VT 1 mA iD = 1 × 10−3 e(v O −0.7)/VT iD v O = 0.7 + VT ln 1 mA
0.8
2
to 55.4
1
1
2
v I = v O + iD × 1 k Using these equations, calculate v I for the different values of v O . For D2 , v I = v O − iD × 1 k
0.8
It is a soft limiter with a gain K 1 and L+ 0.7 V, L− −0.7 V
4.92 (a)
10 k vI
vO
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vI (V)
Chapter 4–38
(b)
VB = 2VD + VC
10 k
and the voltage VA is given by
VA = VB + I × 5k
vO
vI
VA = VB + 5I2
10 k vI
vO
Table 1
4.93
R
1.5 V
vO
1 k
1.5 V
(4)
Equations (1), (2), (3), and (4) can be used to find VB and VC versus VA . We start with a value for VD , use (1) to determine I2 , use (2) to determine VC , use (3) to determine VB , and finally use (4) to determine VA . The results are given in Table 1.
(c)
vI
(3)
In the nonlimiting region 1000 vO = ≥ 0.94 vI 1000 + R
VD3 , VD4 (V)
I2 (mA)
VC (V)
VB = VC +VD3 + VD4
VA (V)
0.4
0
0
0.8
0.8
0.5
0.00003
0
1.0
1.0
0.6
0.002
0.002
1.202
1.212
0.7
0.1
0.1
1.5
2.0
0.73
0.332
0.332
1.792
3.452
0.735
0.406
0.406
1.876
3.91
0.74
0.495
0.495
1.975
4.45
0.745
0.605
0.605
2.095
5.12
For VA < 0, D3 and D4 are cutoff, I2 = 0, VC = 0, and D1 and D2 are conducting a current I1 ,
R ≤ 63.8
I1 = 0.1 e(VD −0.7)/0.025 , mA
(5)
The voltage VB is given by
4.94
5 k VA
VB I
D1 I1
D2
I2
VB = −2VD
(6)
and the voltage VA is
D3
VA = VB − 5I1
D4
Equations (5)–(7) can be used to obtain VB versus VA for negative values of VA . The results are given in Table 2.
VC
(7)
1 k Table 2 VD1 , VD2 (V)
I1 (mA)
VB (V)
VA (V)
When VA > 0, D1 and D2 are cut off and D3 and D4 conduct a current I2 . Since the diodes are 0.1-mA devices, the current I2 is related to the diode voltage VD as follows:
0.4
0
– 0.8
– 0.8
0.5
0
–1.0
–1.0
0.6
0.002
–1.2
–1.21
I2 = 0.1 × e(VD − 0.7)/0.025 , mA
0.7
0.1
–1.4
–1.9
0.73
0.332
–1.46
–3.12
0.74
0.495
–1.48
–3.955
0.75
0.739
–1.5
–5.20
Figure 1
(1)
The voltage VC is given by VC = I2 × 1 k = I2 , V
(2)
where I2 is in mA, and the voltage VB is given by
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Chapter 4–39
Figure 2 shows plots for VB and VC versus VA using the data in Tables 1 and 2. Finally, Figure 3 shows the waveforms obtained at B and C when a 5-V peak, 100-Hz sinusoid is applied at A.
v O = 1 + v D1 + i1 × 1 = 1 + v D1 + i1
(2)
v I = v O + 3i1
(3)
where i1 is in mA and v D1 , v O and v I are in volts. Using these relationships we obtain:
VB, VC (V) VB
2
VC
1
VC 5 4 3 2 1
0
1
2
3
4
1
5 VA (V)
2
Figure 2
5
VA, VB, VC (V)
4 3 2 1 VB
VA VB
i1 (mA)
v D1 (V)
v O (V)
v I (V)
0.01
0.584
1.594
1.625
0.1
0.642
1.742
2.042
0.2
0.660
1.860
2.460
0.5
0.682
2.182
3.682
1.0
0.700
2.700
5.700
1.5
0.710
3.210
7.710
1.8
0.715
3.515
8.915
2.0
0.717
3.717
9.717
For v I < 0, D1 will be cut off and the circuit reduces to that in Fig. 2.
VC
3 k
t
1 2 3 4 5
vI
vO i2
i3
i2
D2
D3
1 k i4
2 V
Figure 3
Figure 2 4.95 Refer to the circuit in Fig. P4.95. For v I > 0, D2 and D3 are cut off, and the circuit reduces to that in Fig. 1.
Here, for v I < −2.5, D2 will be cut off, i2 0 and D3 also will be cut off, and vO = vI
3 k vI
vO i1 i 1
D1 1 k 1 V
Figure 1 Now, for v I < 1.5 V, diode D1 will be off, i1 0, and vO = vI As v I exceeds 1.5 V, diode D1 will turn on and i1 (1) v D1 = 0.7 + 0.025 ln 1
As v I reduces below about −2.5 V, D2 begins to conduct and eventually D3 also conducts. The details of this segment of the v O − v I characteristic can be obtained using the following relationships: i3 v D3 = 0.7 + 0.025 ln 1 v D3 = v D3 i4 = 1 k i2 = i3 + i4 i2 v D2 = 0.7 + 0.025 ln 1 v O = −2 − v D3 − v D2 v I = v O − 3i2
SEDRA-ISM: “P-CH04-086-097” — 2014/11/12 — 16:26 — PAGE 39 — #6
Chapter 4–40
This figure belongs to Problem 4.95, part c.
vO (V)
D1 has an almost constant voltage and the 1 k and the 3 k form a voltage divider
4 3
slope 2 slope 1 All diodes are cutoff
1
10 9 8 7 6 5 4 3 2 1 0 1 D2 and D3 are conducting, 21.4 3.4 V limiting vO slope
1 4
1
2
3
4
5
vt (V) 6
7
8
9 10
2 3
0.013
3.6 Figure 3 The complete transfer characteristic v O versus v I can be plotted using the data in the tables above. The result is displayed in Fig. 3.
In all these equations, currents are in mA and voltages are in volts. The numerical results obtained are as follows: i3
v D3 , i 4
i2
v D2
vO
vI
0
0.01
0.01
0.585
−2.60
−2.63
Slopes: For v I near +10 V the slope is approximately determined by the voltage divider composed of the 1 k and the 3 k,
0
0.05
0.05
0.625
−2.68
−2.85
Slope =
0
0.1
0.1
0.642
−2.74
−3.04
0
0.2
0.2
0.660
−2.86
−3.46
For v I near −10 V, the slope is approximately given by
0
0.3
0.3
0.670
−2.97
−3.87
Slope −
0
0.4
0.4
0.677
−3.08
−4.28
From the table above,
0
0.5
0.5
0.683
−3.18
−4.68
0.01
0.585
0.595
0.687
−3.28
−5.07
0.1
0.642
0.742
0.693
−3.35
−5.56
i2 2.21 mA
0.2
0.660
0.860
0.696
−3.36
−5.94
Thus
0.5
0.682
1.182
0.704
−3.38
−6.93
Slope =
1.0
0.700
1.700
0.713
−3.41
−8.51
1.5
0.710
2.210
0.720
−3.43
−10.06
1 = 0.25 V/V 3+1
rd 2 + rd 3 rd 2 + rd 3 + 3 k
i3 1.5 mA
⇒ ⇒
25 = 16.7 1.5 25 = = 11.3 2.21
rd 3 = rd 2
11.3 + 16.7 = 0.009 V/V 11.3 + 16.7 + 3000
which is reasonably close to the value found from the graph.
SEDRA-ISM: “P-CH04-086-097” — 2014/11/12 — 16:26 — PAGE 40 — #7
Chapter 4–41
At = T1 = T , v O = V1
4.96
= V1 e−T /RC
C
vI
D v O
V1 0
V2
t
V´2
vO
vI 5 2 t
T1
t
0 0
V´1
5 2
T2
where for T CR
V1 V1 (1 − T /CR) = V1 (1 − α) where α 1
From the figure we see that √ v Oav = −5 2 = −7.07 V
During the interval T2 , we have |v O | = |V2 | e−t/(CR/2)
At the end of T2 , t = T , and v O = |V2 |
4.97
where (a)
10 V 10 V
|V2 | = |V2 | e−T /(CR/2) T |V2 | 1 − = |V2 | (1 − 2α) RC/2 Now
V1 + |V2 | = 20 ⇒ V1 + |V2 | − αV 1 = 20
(b)
(1)
20 V
and V2 + V1 = 20 ⇒ V1 + |V2 | − 2α |V2 | = 20 (2)
0V
From (1) and (2) we find that V1 = 2|V2 |
(c)
0V 20 V
Then using (1) and neglecting αV1 yields 3 |V2 | = 20 ⇒ |V2 | = 6.67 V V1 = 13.33 V The result is
13.33 V
(d)
0V 6.67 V 20 V (g)
18 V
(e)
10 V 2 V 10 V
(f) Here there are two different time constants involved. To calculate the output levels, we shall consider the discharge and charge wave forms. During T1 ,
v O = V1 e−t/RC
(h) Using a method similar to that employed for case (f) above, we obtain
13.33 V 6.67 V
SEDRA-ISM: “P-CH04-086-097” — 2014/11/12 — 16:26 — PAGE 41 — #8
